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513. 找树左下角的值

English Version

题目描述

给定一个二叉树，在树的最后一行找到最左边的值。

示例 1:

输入:

    2
   / \
  1   3

输出:
1

 

示例 2:

输入:

        1
       / \
      2   3
     /   / \
    4   5   6
       /
      7

输出:
7

 

注意: 您可以假设树（即给定的根节点）不为 NULL。

解法

“BFS 层次遍历”实现。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findBottomLeftValue(self, root: TreeNode) -> int:
        res = 0
        q = collections.deque([root])
        while q:
            res = q[0].val
            n = len(q)
            for _ in range(n):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return res

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        int res = 0;
        while (!q.isEmpty()) {
            res = q.peek().val;
            for (int i = 0, n = q.size(); i < n; ++i) {
                TreeNode node = q.poll();
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return res;
    }
}

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